To compute the arithmetic mean and the median for the given frequency distribution, let's break down the steps.

Data:

• Class Intervals: $20-30, 30-40, 40-50, 50-60, 60-70$
• Frequencies: $8, 26, 30, 20, 16$

Step 1: Compute the Arithmetic Mean

1. Midpoint (Class Mark) of each class interval:

   • $20-30$: $\frac{20+30}{2} = 25$
   • $30-40$: $\frac{30+40}{2} = 35$
   • $40-50$: $\frac{40+50}{2} = 45$
   • $50-60$: $\frac{50+60}{2} = 55$
   • $60-70$: $\frac{60+70}{2} = 65$

2. Multiply the Midpoint by the Frequency for each class:

   • For $20-30$: $25 \times 8 = 200$
   • For $30-40$: $35 \times 26 = 910$
   • For $40-50$: $45 \times 30 = 1350$
   • For $50-60$: $55 \times 20 = 1100$
   • For $60-70$: $65 \times 16 = 1040$

3. Summation of the midpoints times the frequencies: $200 + 910 + 1350 + 1100 + 1040 = 4600$

4. Summation of the Frequencies: $8 + 26 + 30 + 20 + 16 = 100$

5. The Arithmetic Mean is given by: $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4600}{100} = 46$

Step 2: Compute the Median

1. Find the Cumulative Frequencies:

   • $20-30$: $8$
   • $30-40$: $8 + 26 = 34$
   • $40-50$: $34 + 30 = 64$
   • $50-60$: $64 + 20 = 84$
   • $60-70$: $84 + 16 = 100$

2. Locate the Median Class: The total frequency is $N = 100$. The median is located at the $\frac{N}{2} = \frac{100}{2} = 50$-th observation. The cumulative frequency just greater than 50 is $64$, so the median class is $40-50$.

3. Apply the Median Formula: $\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f_m} \right) \times h$ Where:

   • $L = 40$ (lower boundary of the median class)
   • $F = 34$ (cumulative frequency of the class before the median class)
   • $f_m = 30$ (frequency of the median class)
   • $h = 10$ (class width)

   Substituting the values:

   = 40 + \left( \frac{16}{30} \right) \times 10 = 40 + 5.33 = 45.33$$

Summary:

• Arithmetic Mean: $46$
• Median: $45.33$

Let me know if you need further clarifications or have any questions.

Here are some related questions:

1. How would the mean change if we increased the frequency of the last class interval?
2. Can you explain why the median is less than the mean in this case?
3. How would you calculate the mode for this data set?
4. What happens if the class intervals are unequal? How does that affect the calculation?
5. Could we use graphical methods like histograms to estimate the median or mean?

Tip: Always ensure that class intervals are continuous for accurate calculations of median and mean in grouped data.